What is the solution for 3/4*x^{''}+36x=0,x^'(0)=0,x(0)=-6 ?

The solution for 3/4*x^{''}+36x=0,x^'(0)=0,x(0)=-6 is x=-6cos(4sqrt(3)t)

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3/4*x^{''}+36x=0,x^'(0)=0,x(0)=-6

\frac{3}{4} \cdot x^{^{''}} + 36 x = 0, x^{^{'}} (0) = 0, x (0) = - 6

x = - 6 cos (43 t)

Solution steps

\frac{3}{4} \cdot x^{''} (t) + 36 x = 0

Solve linear ODE:

x = - 6 cos (43 t)

x = - 6 cos (43 t)

What is the solution for 3/4*x^{''}+36x=0,x^'(0)=0,x(0)=-6 ?
The solution for 3/4*x^{''}+36x=0,x^'(0)=0,x(0)=-6 is x=-6cos(4sqrt(3)t)